是否可以将forward-declared类的member函数声明为friend？我正在努力做以下事情：
class BigComplicatedClass;
class Storage {
int data_;
public:
int data() { return data_; }
// OK,but provides too broad access:
friend class BigComplicatedClass;
// ERROR "invalid use of incomplete type":
friend void BigComplicatedClass::ModifyStorage();
};
所以目标是(i)将朋友的声明限制在一个单一的方法中,(ii)不要包含复杂类的定义来减少编译时间.
// In Storage.h:
class BigComplicatedClass_Helper;
class Storage {
// (...)
friend class BigComplicatedClass_Helper;
};
// In BigComplicatedClass.h:
class BigComplicatedClass_Helper {
static int &AccessData(Storage &storage) { return storage.data_; }
friend void BigComplicatedClass::ModifyStorage();
};
然而,这似乎有点笨拙…所以我认为必须有一个更好的解决方案！
解决方法
正如@Ben所说,这是不可能的,但是您可以通过
“passkey”给该成员函数提供特定的访问.它有点像中间助手类,但是更清晰：
// Storage.h
// forward declare the passkey
class StorageDataKey;
class Storage {
int data_;
public:
int data() { return data_; }
// only functions that can pass the key to this function have access
// and get the data as a reference
int& data(StorageDataKey const&){ return data_; }
};
// BigComplicatedClass.cpp
#include "BigComplicatedClass.h"
#include "Storage.h"
// define the passkey
class StorageDataKey{
StorageDataKey(){} // default ctor private
StorageDataKey(const StorageDataKey&){} // copy ctor private
// grant access to one method
friend void BigComplicatedClass::ModifyStorage();
};
void BigComplicatedClass::ModifyStorage(){
int& data = storage_.data(StorageDataKey());
// ...
}
